Q63 / 75 · JEE Main 2026 (06 April Shift 2) · Physics
A current of $30$ A each flows in opposite directions in two conducting wires, placed parallel to each other at a distance of $8$ cm. The magnetic field at the mid point between the two wires is __________ $\mu$T. $\left(\dfrac{\mu_0}{4\pi} = 10^{-7}\text{ N/A}^2\right)$
- A$30$
- Correct answerB$300$
- C$150$
- D$0.0$
Solution
Let the distance between the wires be $d = 8\text{ cm}$. The distance of the midpoint from each wire is $r = \dfrac{d}{2} = 4\text{ cm} = 0.04\text{ m}$.
The current in each wire is $I = 30\text{ A}$.
Since the currents flow in opposite directions, the magnetic fields produced by both wires at the midpoint point in the same direction and add up.
The total magnetic field at the midpoint is:
$B_{\text{net}} = B_1 + B_2 = \dfrac{\mu_0 I}{2\pi r} + \dfrac{\mu_0 I}{2\pi r} = \dfrac{\mu_0 I}{\pi r}$
Substituting the given values:
$B_{\text{net}} = \dfrac{4\pi \times 10^{-7} \times 30}{\pi \times 0.04}$
$B_{\text{net}} = \dfrac{120 \times 10^{-7}}{0.04}$
$B_{\text{net}} = 3000 \times 10^{-7}\text{ T}$
$B_{\text{net}} = 300 \times 10^{-6}\text{ T} = 300\ \mu\text{T}$
Answer: $300$
The current in each wire is $I = 30\text{ A}$.
Since the currents flow in opposite directions, the magnetic fields produced by both wires at the midpoint point in the same direction and add up.
The total magnetic field at the midpoint is:
$B_{\text{net}} = B_1 + B_2 = \dfrac{\mu_0 I}{2\pi r} + \dfrac{\mu_0 I}{2\pi r} = \dfrac{\mu_0 I}{\pi r}$
Substituting the given values:
$B_{\text{net}} = \dfrac{4\pi \times 10^{-7} \times 30}{\pi \times 0.04}$
$B_{\text{net}} = \dfrac{120 \times 10^{-7}}{0.04}$
$B_{\text{net}} = 3000 \times 10^{-7}\text{ T}$
$B_{\text{net}} = 300 \times 10^{-6}\text{ T} = 300\ \mu\text{T}$
Answer: $300$
Hence, the correct answer is option: 'B'.